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I am often confused by the plethora of formulations of this result, which invariably differ from text to text. Sometimes it's explicitly about reducibility of polynomials, sometimes it's about content and its multiplicativity. Often it's about $\Bbb Z$ and $\Bbb Q$, but sometimes it's more general. Is there any way of stating this result in maximal generality and clarity? My understanding of it is the following:

Let $D$ be a UFD and $K$ its field of fractions. Then:

If $f\in D[x]$ is irreducible over $D$, it is irreducible over $K$.

And, (somewhat) equivalently:

If $f\in K[x]$ is reducible, i.e. $f=gh$, then $\exists \overline{g},\overline{h}\in D[x]$ such that $f=\overline{gh}$.

That is, moving from a UFD into its field of fractions doesn't affect irreducibility and vice versa for reducibility. Is this a valid interpretation? If so, why is it important to understand the primitive part/content aspect of this lemma? Is it possible to be any more general than this (assuming it's correct)?

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Hilbert Jr.
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    The proof of the result relies on an argument about content, and that the product of primitive polynomials is primitive; though Gauss did not phrase it this way (relying instead on an analysis of prime powers in the coefficients of the factors), it greatly simplifies the arguments by making them conceptual instead of calculation-intensive. See here and the post linked to in the comment. – Arturo Magidin Jun 07 '20 at 22:47
  • See this formulation. A UFD is integrally closed in its field of fractions. The result holds generally for monic polynomials over an integrally closed subring of a commutative ring (not necessarily a domain). – Matthé van der Lee Jun 16 '20 at 14:29

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